Special case when solving D.E.'s

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  • #1
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Why can't there be common terms on both complementary function and particular integral when solving differential equations?

For instance,

dy/dx + 3y = exp(-x) + exp(-3x)

y(CF) = Aexp(−3x)

y(PI) = Cexp(−x) + Dxexp(−3x)

The term Dexp(-3x) in the P.I. has to be multiplied by x to be linearly independent of Aexp(-3x) in the C.F.. Why? What would happen if it wasn't?
 

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  • #2
HallsofIvy
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Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.
 
  • #3
arildno
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Note that due to the characteristics of the differentiation of a product, ALL products of the form Y(x)=f(x)e^(-3x) will experience some "internal cancellation".

In your case, we will get on LHS:
[tex]\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}[/tex]
Thus, since this must equal identically RHS, i.e, [tex]e^{-3x}[/tex], we get the subsidiary equation on f:
[tex]f^{,}(x)=1[/tex], which is readily solved.
 
  • #4
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Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.
D would vanish... That's why they have to be linearly independent? I can't think of it in practical situations.
 
  • #5
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[tex]\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}[/tex]
Where did you get this from?
 
  • #6
arildno
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From defining Y(x) as f(x)e^(-3x), and applying the LHS operator on it!
 

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